−1 ⎛ b ⎞
−1 ⎛ −b ⎞
So what's up? Is φ = tan ⎜ ⎟ or is φ = tan ⎜ ⎟ ? To clear up this confusion, let's do some
⎝ a⎠
⎝ a ⎠
forensics.
⎛ −b ⎞
Back in slide 19-4 we solved for θ = tan −1 ⎜ ⎟ but here we were after something slightly
⎝ a ⎠
different. Specifically,
cos(θ + ωt ) = cos(θ ) cos(ωt ) − sin (θ ) sin(ωt ) = acos(ω t ) + bsin(ωt )
so in this case, a = cos(θ ) , b = − sin(θ ) , then, we use the fact that tan x= sin x/cos x. The
negative sign arises because of the negative sign in the formula when taking the cos of the sum
of two angles (i.e., b=-sin()).
Later, in lecture 20 (and onward), we had,
Hy
Op
Φ
Ad
In general, from trigonometry we know:
Op
Ad
Op
sin(φ ) =
; cos(φ ) =
; tan(φ ) =
Hy
Hy
Ad
So going from Cartesian form, (Ad, Op), to polar form, (Hy, Φ):
⎛ Op⎞
φ = tan −1 ⎜ ⎟ ; Hy = Op2 + Ad 2
⎝ Ad ⎠
Unlike lecture 19, this time there is no negative sign in the tan-1
Going from polar form to Cartesian form:
Op = Hy *sin (φ ) ; Ad = Hy * cos(φ )